A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio
1 answer:
Explanation:
It is given that,
Weight of the rock in air, W = 110 N
Since, W = mg
m = 11.22 kg
We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.
d is the density of water,
V is the volume of rock,
M = 3.37 kg
The apparent weight in water, W = m - M
W = 76.93 N
So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s
Answer:
2.1 rad/s
Explanation:
Given that,
Mass of a tether ball, m = 0.546 kg
Length of a rope, l = 4.56 m
The maximum tension the rope can withstand before breaking is 11.0 N
We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :
Let is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :
So, the maximum angular speed of the ball is 2.1 rad/s.